Question

The value of $f(0)$ so that the function $f(x)=\frac{1-\cos \left(1-\cos x\right)}{x^4}$ is continuous everywhere is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{6}$
• D. $\frac{1}{8}$

Answer 1

Answer: (D)

$\underset{x\to 0}{\lim }\frac{1-\cos (1-\cos x)}{x^4}$
$=\underset{x\to 0}{\lim }\frac{2{\sin }^2\left(\frac{2{\sin }^2\left(\frac{x}{2}\right)}{2}\right)}{x^4}$
$=2\underset{x\to 0}{\lim }\frac{{\sin }^2\left({\sin }^2\left(\frac{x}{2}\right)\right){\left({\sin }^2\left(\frac{x}{2}\right)\right)}^2}{x^4{\left({\sin }^2\left(\frac{x}{2}\right)\right)}^2}$
$=2\underset{x\to 0}{\lim }\frac{{\sin }^4\left(\frac{x}{2}\right)}{{\left(\frac{x}{2}\right)}^4{2}^4}=\frac{1}{2^3}=\frac{1}{8}$